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Forum:Can you help me answer my 12 year old daughters googolology question?
Hi, Recently my 12 year old daughter was asking me about big numbers. I've always had a fascination myself - but am very much a layman by the standards of people on this site. She knew about googolplex and wanted to know if there were bigger named numbers. I told her grahams number was much much bigger and tried to explain the construction. She got a bit lost with the multiple up-arrow notation but I wanted to encourage her interest so I set her a challenge - if she could describe a way to construct a number bigger than grahams number I would buy her a new phone! This seemed a safe bet for me as I don't think its an easy task from where she is starting - but it got her interest. Below is what she came up (she did it with pictures and hand waving - not this notation - but I've tried to formalise it for her) : Her first idea was power towers, ie her first number was power tower of googolplexes, googolplex high, ie : A = googolplex ↑↑ googolplex I explained this didn't even begin to get to the first step of grahams number. She then tried: : B = A ↑↑ A Again I explained that wasn't big enough. Then she came up with a way of iterating the process : : f(0) = B : f(n) = f(n-1) ↑↑ f(n-1) and her third number was: : C = f(B) I said it still wasn't enough (though she was getting further than I'd imagined and I realised I was going to have trouble justifying that!). Finally she iterated again: : D = f(f(f(f(... f©)...) with C nested f's. I told her I still didn't think it was big enough - but I wasn't so confident. I said it was probably bigger than G1 (3 ↑↑↑↑ 3) but still nowhere near G64? Can anyone help me understand how far she got? I need to be able to justify to her that I don't owe her a new phone! thanks - reddal :: Her cool number is bounded by \(G_2\) in the following way: \begin{eqnarray*} (n+2) \uparrow^2 (n+2) & = & (n+2) \uparrow^3 2 < 2 \uparrow^3 (n+2) \\ A & = & 10^{10^{100}} < (2^4)^{(2^4)^{(2^4)^2}} = 2^{2^{2^{10}+2}} < 2^{2^{2^{2^{2^2}}}} = 2 \uparrow^2 6 \\ & < & 6 \uparrow^2 6 < 2 \uparrow^3 6 \\ B & = & A \uparrow^2 A < 2 \uparrow^3 (2 \uparrow^3 6) = 2 \uparrow^3 7 \\ f(0) & = & B < 2 \uparrow^3 7 \\ f(1) & = & f(0) \uparrow^2 f(0) < 2 \uparrow^3 (2 \uparrow^3 7) = 2 \uparrow^3 8 \\ \vdots & \vdots & \vdots \\ f(n) & < & 2 \uparrow^3 (n+7) \\ C & = & f(B) < 2 \uparrow^3 ((2 \uparrow^3 7)+7) < 7 \uparrow^3 (7 \uparrow^3 7) = 7 \uparrow^4 3 \\ f© & < & 2 \uparrow^3 ((7 \uparrow^4 3) + 7) < 7 \uparrow^3 (7 \uparrow^3 (7 \uparrow^4 3)) = 7 \uparrow^4 5 \\ \vdots & \vdots & \vdots \\ f^n© & < & 7 \uparrow^4 (n+3) \\ D & = & f^C© < 7 \uparrow^4 ((7 \uparrow^4 3)+3) < 7 \uparrow^4 (7 \uparrow^4 (7 \uparrow^4 7)) = 7 \uparrow^5 4 \\ & < & (3 \uparrow 3) \uparrow^5 (3 \uparrow 3) \ll 3 \uparrow^{G_1} 3 = G_2 \end{eqnarray*} :: p-adic 15:33, October 5, 2018 (UTC) Fantastic! Thanks so much. - reddal